![]() ![]() No need to create the list in memory if you just want to push it to a file. Just use itertools.permutations and write the objects to the file on-the-fly. ![]() Second, if you just want to "save" the items to some kind of massive file, you do NOT need to realize the list in memory. Given that I'm not too sure what you would do with a saved list of 6B things, but I think you have 2 strategies if you want to go forward.įirst, you could reduce the size of the things in the list by substituting something like a numpy unit8 for each item, which would reduce the size of the resulting list by a LOT, but you would not have the format you want. How can I make this code more memory efficient so that it doesn't crash for a large list? If I need to, I can run the code separately for each item separator (commas, periods, directly joined). Names_t = list(set(names_t)) # remove all duplicates Perms = filter(None, perms) # remove empty lists, to prevent two symbols in a row or a symbol before/after Perms = # remove empty nested elements, to prevent two symbols in a row or a symbol before/after Even when using a machine with a large amount of RAM. But then, this fails with a MemoryError when removing empty elements after converting to a list. I've used itertools.permutations to create an itertools.chain ( perms). # Also creating groups of length 1, 2, and 3 # Avoiding a symbol at the beginning or end of a resulting string for all the possible groups of 4, order is important # ideal output would be a list that contains strings like the following:Īpple-banana-bread (no dashes before or after!)Īpple-banana (by combining with the first item in the list, I also get shorter groups but need to delete empty items before joining) The goal is to pair items in groups of four or less, separated by 1) periods, 2) dashes, and 3) without any separation. However, the itertool.permutations option is most suitable.I'm trying to run a very large permutation using Python. There were two alternative things that you can do: We will be using the combinations () and permutations () methods under the itertools module of Python. We will look at sets of characters and numbers. The result is a big list: [('PAN', 'AMS', 'CAS', 'NYC', 'HEL')Īs a side note, whilst your attempt was good, because the output is large and the number of recursions is large, you could get stack overflow. In this tutorial, we will learn how to get the permutations and combinations of a group of elements in Python. ![]() # or this is you want to inspect the loop You can user the itertools module for this.īasically, when doing permutations and combinations, this is a very useful place. We use permutations with your portnames data now - portnames = Print(f"permutation ")īreak # <- stops generating additional permutations If abs(p.index("□") - p.index("□")) = 1:Īdditionally generators can be paused/stopped at any time, allowing us to skip potentially millions of computations for significantly large problems - # which permutation is in rainbow order?įor (n, p) in enumerate(permutations(t)): ![]() # find all permutations where blue and yellow are adjacent Generators are a good fit for combinatorics because working with them is natural and straightforward - # find all permutations where red is left of green Yield from map(lambda r: (t, *r), inserts(t, x)) # <. We can simplify the above definitions while maintaining the exact same behaviour - def permutations(t): Python has strong support for generators and offers delegation using yield from. yield x prepended to t and for all i of the recursive sub-problem inserts(t, x) yield t prepended to i (inductive) t has at least one element.If the input t is empty, yield the final insert, (x).Where inserts(t, x) can be written using inductive reasoning as well. For all p of the recursive sub-problem permutations(t), for all i of inserts(p, t), yield iįor p in permutations(t): # 2. if t is empty, yield the empty permutation, ().We can write permutations(t) of any iterable t using inductive reasoning. ![]()
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